Wait until service starts in bash-script

Question

I've a bash-script that starts some service in background. After this service successfully starts it prints "Server is active" to the stdout. I need to wait until this string appears and then continue executing my script. How can I achieve this?

Answer 1

I would do in this way.

./server > /tmp/server-log.txt &
sleep 1
while ! grep -m1 'Server is active' < /tmp/server-log.txt; do
    sleep 1
done

echo Continue

Here -m1 tells grep(1) to quit at the first match.

I veryfied my answer with my toy "service" below:

#! /bin/bash

trap "echo 'YOU killed me with SIGPIPE!' 1>&2 " SIGPIPE

rm -f /tmp/server-output.txt
for (( i=0; i<5; ++i )); do
    echo "i==$i"
    sleep 1;
done
echo "Server is active"
for (( ; i<10; ++i )); do
    echo "i==$i"
    sleep 1;
done
echo "Server is shutting down..." > /tmp/server-output.txt

If you replace echo Continue with echo Continue; sleep 1; ls /tmp/server-msg.txt, you will see ls: cannot access /tmp/server-output.txt: No such file or directory which proves the "Continue" action was triggered right after the output of Server is active.